Index shifting in summation FORMULAS

So I am trying to derive a formula from one of the standard summation formulas except starting at a different index. So if I have the series..

[tex]\sum i = \frac{n(n+1)}{2}[/tex]

Where "i" runs from 1 to n. (I dont know how to put it in the code.) If I want to make the series start from zero, I would use another index equal to, "k=i-1" And now my upper limit will be "n-1" So now I can replace "i" in the sum with "k+1", Shouldn't the new formula for the partial sums be..

The first sum is the same whether it starts with [itex]k=0[/itex] or [itex]k=1[/itex], but the second sum is not.

Your mistake is in your calculation of the second sum. How many 1's are being added?

Hmm I think I still have some issues. Since the original formula for the partial sums is..

[tex]\sum_{i=1}^{n} i = \frac{n(n+1)}{2}[/tex]

Since "n" represents the upper limit, the new formula for the series I am after should be just replacing all the "n" with "n-1."
[tex]\sum_{k=0}^{n-1} k+1 =\frac{(n-1)(n)}{2} + \sum_{k=0}^{n-1} 1[/tex]

Now you say up until this part is correct? For the second sum, shouldn't the sum of this be just the upper limit? For example..

Since "n" represents the upper limit, the new formula for the series I am after should be just replacing all the "n" with "n-1."
[tex]\sum_{k=0}^{n-1} k+1 =\frac{(n-1)(n)}{2} + \sum_{k=0}^{n-1} 1[/tex]

Actually both the lower and the upper limit have changed, right? You are now summing from 0 to n-1, whereas the original formula sums from 1 to n.

Now it happens that

[tex]\sum_{k=0}^{n-1}k = \sum_{k=1}^{n-1}k[/tex]

i.e. it doesn't matter whether you start summing from [itex]k=0[/itex] or [itex]k=1[/itex] in THIS PARTICULAR SUM. That is because the summand is [itex]k[/itex], so the [itex]k=0[/itex] term is 0: it has no effect on the value of the sum. You can therefore start summing from [itex]k=1[/itex] and get the same answer.

Now you say up until this part is correct?

Yes, it's correct up to this point.

For the second sum, shouldn't the sum of this be just the upper limit? For example..

The summand is [itex]1[/itex], and you are summing it [itex]n[/itex] times. The [itex]k=0[/itex] term DOES matter in this case, because that term is not zero.

Therefore

[tex]\sum_{k=0}^{n-1} 1 = n[/tex]

Think of it as having [itex]n[/itex] marbles with labels on them. It doesn't matter if the labels read [itex]0[/itex] through [itex]n-1[/itex] or [itex]1[/itex] through [itex]n[/itex], there are still [itex]n[/itex] marbles.

Yes, it looks right to me. And I tried plugging in n = 20 and verified that the two formulas give the same answer. Of course they're really the same formula, in the sense that if you carry out the addition of the three terms in your expression by putting them over a common denominator and simplifying, you should end up with the original formula.